Question

Bad gums may mean a bad heart. Researchers discovered that 79% of people who have suffered a heart attack had periodontal disease, an inflammation of the gums. Only 33% of healthy people (those who have not had heart attacks) have this disease. Suppose that in a certain community heart attacks are quite rare, occurring with only 15% probability.
A. If someone has periodontal disease, what is the probability that he or she will have a heart attack?
B. If 38% of the people in a community will have a heart attack, what is the probability that a person with periodontal disease will have a heart attack?

Answer 1

Answer:

(A) 0.297

(B) 0.595

Step-by-step explanation:

Let,

H = a person who suffered from a heart attack

G = a person has the periodontal disease.

Given:

P (G|H) = 0.79, P(G|H') = 0.33 and P (H) = 0.15

Compute the probability that a person has the periodontal disease as follows:

$P(G)=P(G|H)P(H)+P(G|H')P(H')\\=P(G|H)P(H)+P(G|H')(1-P(H))\\=(0.79\times0.15)+(0.33\times(1-0.15))\\=0.399$

(A)

The probability that a person had periodontal disease, what is the probability that he or she will have a heart attack is:

$P(H|G)=\frac{P(G|H)P(H)}{P(G)}\\=\frac{0.79\times0.15}{0.399} \\=0.29699\\\approx0.297$

Thus, the probability that a person had periodontal disease, what is the probability that he or she will have a heart attack is 0.297.

(B)

Now if the probability of a person having a heart attack is, P (H) = 0.38.

Compute the probability that a person has the periodontal disease as follows:

$P(G)=P(G|H)P(H)+P(G|H')P(H')\\=P(G|H)P(H)+P(G|H')(1-P(H))\\=(0.79\times0.38)+(0.33\times(1-0.38))\\=0.5048$

Compute the probability of a person having a heart attack given that he or she has the disease:

$P(H|G)=\frac{P(G|H)P(H)}{P(G)}\\=\frac{0.79\times0.38}{0.5048}\\ =0.59469\\\approx0.595$

The probability of a person having a heart attack given that he or she has the disease is 0.595.